Driving with a dog cause you just have to mixed them to ether that’s it
Answer:
a) See attachment 1.
b) t² and D
c) See attachment 2.
d) g = 9.8 m/s² (1 d.p.)
Step-by-step explanation:
<h3><u>Part (a)</u></h3>
See attachment 1. The line of best fit is shown in red.
<h3><u>Part (b)</u></h3>
The quantities the student should graph in order to produce a <u>linear relationship</u> between the two quantities are t² and D.
<h3><u>Part (c)</u></h3>
Make a table of values of t² and D:

<u>Plot</u> a graph of D against t² and draw a line of best fit (see attachment 2).
<h3><u>Part (d)</u></h3>
From inspection of the graph, the line of best fit passes through the origin (0, 0) and (0.1024, 5.0). Therefore, use these two points to find the slope of the line:

Therefore:



The statement that best describes the relationship between lines s and t is: Lines s and t are parallel.
<h3>What are parallel lines?</h3>
Parallel lines can be defined as those lines that are equal to one another but does not intersect one another.
For Lines s and t to be parallel lines this means that lines s ant t are cut by transversal. Hence, transversal intersect lines s and t to make it parallels lines.
Therefore Lines s and t are parallel.
Learn more about parallel lines here:brainly.com/question/24607467
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<span>Simplifying
3x + 10 = 5x + 22
Reorder the terms:
10 + 3x = 5x + 22
Reorder the terms:
10 + 3x = 22 + 5x
Solving
10 + 3x = 22 + 5x
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Add '-5x' to each side of the equation.
10 + 3x + -5x = 22 + 5x + -5x
Combine like terms: 3x + -5x = -2x
10 + -2x = 22 + 5x + -5x
Combine like terms: 5x + -5x = 0
10 + -2x = 22 + 0
10 + -2x = 22
Add '-10' to each side of the equation.
10 + -10 + -2x = 22 + -10
Combine like terms: 10 + -10 = 0
0 + -2x = 22 + -10
-2x = 22 + -10
Combine like terms: 22 + -10 = 12
-2x = 12
Divide each side by '-2'.
x = -6
Simplifying
x = -6</span>
Answer:
A. x
Step-by-step explanation:
1. Simplify parentheses

2. Divide by 13

3. Basic math (
)
